Answers to Problem Set Number 9 for 18.04.
MIT (Fall 1999)
Rodolfo R. Rosales
Boris Schlittgen
y
z
Zhaohui Zhang.
December 6, 1999
Contents
1 Problems from the book by Sa and Snider.
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
Problem 06 in section 6.3.
Problem 11 in section 6.3.
Problem 01 in section 6.4.
Problem 03 in section 6.4.
Problem 03 in section 6.5.
Problem 10 in section 6.5.
Problem 01 in section 6.6.
Problem 04 in section 6.6.
Problem 06 in section 6.7.
Problem 10 in section 6.7.
Problem 04 in section 7.1.
Problem 07 in section 7.2.
Problem 13 in section 7.2.
Problem 05 in section 7.3.
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MIT,
Department of Mathematics, room 2-337, Cambridge, MA 02139.
y MIT,
Department of Mathematics, room 2-490, Cambridge, MA 02139.
z MIT,
Department of Mathematics, room 2-229, Cambridge, MA 02139.
1
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2
2
3
4
4
5
6
8
9
9
10
11
11
15
Answers to Problem Set # 9.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
2
1 Problems from the book by Sa and Snider.
1.1
Problem 06 in section 6.3.
We want to calculate the integral
I=
1
Z
0
x2
dx ;
(x2 + 1)(x2 + 4)
which we can recast into the form
1Z 1
x2
1Z
dx
=
lim
R!1 2
2 1 (x2 + 1)(x2 + 4)
x2
dx ;
R (x2 + 1)(x2 + 4)
where we have used that the integrand is an even function of x. We now turn the second integral
(the one with the limit R ! 1) into a contour integral in the complex plane, by closing the contour
in the upper half plane with a semicircle of radius R. The contribution to the integral from the
semicircle vanishes as R ! 1, since the degree of the denominator is 2 higher than the degree of
z2
. Then, using residues:
the numerator. Thus, let f (z ) =
(z + i)(z i)(z + 2i)(z 2i)
2i
I =
(Res(f; z = i) + Res(f; z = 2i))
2
!
1
4
= i
+
(2i)(3i)( i) (3i)(i)(4i)
=
6
I=
1.2
R
Problem 11 in section 6.3.
Here we are interested in the integral
I=
1 dx
Z
0
1 + x3
:
In order to do this integral, we consider the integral of f (z ) = 1=(1 + z 3 ) over the path
the boundary of the wedge
SR = z =
and then let R ! 1.
rei
:0
given by
2
and 0 r R
3
2
When R ! 1 the contribution to the integral over by the arc: z = Rei (0 ) vanishes,
3
since the degree of the denominator is 3 higher than the degree of the numerator. Thus only the
Answers to Problem Set # 9.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
3
contributions from the two straight pieces in remain. As R ! 1, one of these pieces becomes the
positive real axis, while the other can be parametrised by z (x) = xe2i=3 , where 1 > x 0 (notice
that this is backwards in x, thus a minus sign will appear.
1
Since f (xe2i=3 ) = f (x) and f (z ) =
, we have
i=
3
(z e )(z + 1)(z e5i=3 )
lim
R!1
Z
f (z )dz =
Z
0
1
f (x)dx +
Z
0
1
f (xe2i=3 )e2i=3 dx = (1 e2i=3 )
Z
1
0
f (x)dx ;
and (using the residue theorem)
lim
R!1
Z
f (z )dz = 2iRes(f; z = ei=3 ) =
(ei=3
2i
+ 1)(ei=3
ei5=3 )
:
Hence,
I=
1.3
1
Z
0
2i
ei5=3 )(1 e2i=3 )
2i
= i=3 i=6 i=6
e e (e + e i=6 )(ei=3 e i=3 )2
2
p
=
=
2
2(2i)2 cos(=6) sin (=3) 4( 3=2)(3=4)
p
2 3
:
=
9
f (x)dx =
(ei=3
+ 1)(ei=3
Problem 01 in section 6.4.
We have
I=
Z
"
Z 1
1 cos(2x)
e2ix
=
Re
1 x2 + 1
1 (x + i)(x
#
i)
"
= Rlim
Re
!1
Z
#
e2ix
:
R (x + i)(x i)
R
We now close the contour (in the real line, from R to R) by a semicircle of radius R in the upper
half plane. The contribution from this semicircle vanishes as R ! 1 (by Jordan's lemma). Thus,
from the residue theorem:
"
e2iz
I = Re 2iRes
(z + i)(z
#
"
2
e
= Re 2i
2i
= 2:
e
!#
i)
;z = i
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
Answers to Problem Set # 9.
4
1 cos(2x)
. As
1 x2 + 1
far as I can tell, this is superuous. This integral converges absolutely; there is no \principal value"
Remark 1.3.1 The book puts a \principal value" label in front of the integral I =
Z
involved:
A principal value label only makes sense for integrals that do not converge absolutely, in which case
more than one meaning can be given to the answer (thus one must clarify which one is being used).
Principal value usually means that the integral is being given a meaning by taking a symmetric limit
near the regions producing the failure of absolute convergence.
1.4
Problem 03 in section 6.4.
We have
I=
Z
"
Z
1 Z 1 cos(x)
1
dx
=
dx
=
lim
Re
R!1 2
(x2 + 1)2
2 1 (x2 + 1)2
1 cos(x)
0
#
eix
dx ;
R (x2 + 1)2
R
eiz
where we have used that the integrand is an even function of x. Now write f (z ) =
(z + i)2 (z i)2
and close the contour of integration by a semicircle of radius R in the upper half plane. The
contribution to the integral from the semicircle vanishes as R ! 1, due to Jordan's Lemma. We
then have
I =
=
=
=
=
1.5
1
Re [2iRes (f (z ); z = i)]
2 "
!#
1
d
eiz
Re 2i lim
z !i dz (z + i)2
2
"
#
1
(z + i)2 ieiz eiz 2(z + i)
Re 2i lim
z !i
2
(z + i)4
1
Re(2i=(2ie))
2
2e
Problem 03 in section 6.5.
Consider now
1
I = P.V.
1 (x
Z
Z
1 Z 2 Z 1
eix
eix
dx = lim
+
+
dx :
!0
1)(x 2)
1
1+
2+ (x 1)(x 2)
Answers to Problem Set # 9.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
5
Note that this integral only makes sense as a principal value (as indicated above) because of the
two poles on the real axis. We can introduce now one more limit1 and write:
I = Rlim
lim
!1 !0
Z
1 R
+
2 Z
1+
+
R
Z
2+
!
eix
dx :
(x 1)(x 2)
We now close the contour of integration using three semicircles in the upper half plane. One large (of
radius R) centered at z = 0 and two small ones (of radius ) centered at z = 1 and z = 2 respectively.
The resulting contour is similar to the one in gure 6.16 (section 6.5, page 278) of the book. Then:
As R ! 1 the contribution from the large semicircle vanishes (by Jordan's lemma).
As ! 0 each one of the small semicircles picks up a contribution of i times the correeiz
sponding residue of f (z ) =
(by Lemma 4, section 6.3, page 275 in the book).
(z 1)(z 2)
Because f has no singularities inside the closed contour, its integral over it vanishes.
Putting all these facts together, we obtain:
I = i (Res(f; z = 1) + Res(f; z = 2))
= i ( ei + e2i )
1.6
Problem 10 in section 6.5.
We consider now the integral
we can write:
1 sin2 (x)
Z
0
x2
dx. Note that the integrand is an even function of x, thus
1 Z 1 sin2 (x)
1 Z 1 12 (1
dx
=
x2
2 1 x2
2 1
0
Z
Z
1 1 cos(2x)
1
= lim
+
dx
!0 4
x2
1 "
#
Z
Z 1 1 e2ix
1
= lim
Re
+
dx = Rlim
lim
!0 4
!1 !0
x2
1 I =
Z
1 sin2 (x)
dx =
cos(2x))
dx
x2
1
Re
4
"
Z
R
+
Z
R
!
1
e2ix
x2
#
dx :
Remark 1.6.1 Notice that the introduction of the limit ! 0 in the second line is necessary because
(although (1 cos(2x))=x2 is not singular at x = 0) once the integral is transformed to the real part
of a complex integral, a singularity at x = 0 appears.
1 The
integral dening I has no convergence problems at x = 1, this extra limit is for the purpose of being able
to use residues.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
Answers to Problem Set # 9.
6
We now close the contour in the upper half plane with a large semicircle (of radius R) centered at
z = 0 and a small semicircle (of radius ) also centered at z = 0. The resulting closed contour is
very similar to the one shown in gure 6.15 (section 6.5, page 277) of the book. Now:
As R ! 1 the contribution from the large semicircle vanishes.
As ! 0 the contribution from the small semicircle picks up a contribution of i times the
1 e2iz
residue at z = 0 of f (z ) =
(by Lemma 4, section 6.3, page 275 in the book | note
z2
that f has a simple pole at z = 0).
Because f has no singularities inside the closed contour, its integral over it vanishes.
Putting it all together, we have:
1
i Res (f (z ); z = 0)
4
d
i
lim
(1 e2iz )
=
z
!
0
4
dz
i
=
lim
( 2ie2iz )
z
!
0
4
= :
2
I =
1.7
Problem 01 in section 6.6.
Here we consider the integral:
I=
1
Z
px
dx ;
1 + x2
where we take the positive square root of x. Take now the branch of the square root function (with
branch cut along the positive real axis) dened by
0
pz = prei ;
pz
1
2
for z = rei , with 0 2. Let now f (z ) =
and consider the integral of f over the \key1 + z2
hole" contour given by2 :
Start on x = and follow the real line, remaining on the Im(z ) > 0 side, all the way up to
x = R. That is, we have z = x + i0, with x R.
2 See
gure 6.21 (section 6.6, page 284) in the book
Answers to Problem Set # 9.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
7
Continue with the circle z = Rei , 0 2.
Continue now moving on the real line, remaining on the Im(z ) < 0 side, all the way up from
x = R back to x = . That is, we have z = x i0, with R x .
Finish the contour with the circle z = ei , 2 0.
Now we have, as R ! 1 and ! 0:
The contribution from the circle of radius R vanishes, since the degree of the denominator
exceeds that of the numerator by 1:5 > 1.
The contribution from the circle of radius vanishes, since the integrand is bounded near
z = 0 and the length of the contour vanishes.
The contribution from the straight line piece on Im(z ) > 0 has I as its limit.
The contribution from the straight line piece on Im(z ) < 0 has the limit:
p
p
Z 1
xe2i
x
i
dx = e
dx = I :
2
0 1 + x2
1 1+x
Z
0
i=2
3i=2
The contour encloses two
pzsingularities (simple poles) at z = i = e and z = i = e .
Thus, since f (z ) =
, we have:
(z + i)(z i)
Z
f (z )dz = 2i(Res(f; z = i) + Res(f; z = i))
!
=
=
=
=
Putting all this together, we get:
ei=4 e3i=4
2i
+
2i
2i
!
ei=4 e i=4
2i
+
2i
2i
2 cos
4
2p :
2
I=p :
2
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
1.8
Answers to Problem Set # 9.
8
Problem 04 in section 6.6.
Here we consider the integral:
x
;
0 (1 + x2 )2
where 1 < < 3, 6= 1 and we take the real positive value for x . We proceed as in the solution
to problem 01 in this section and dene a branch for z (with the branch cut along the positive
real axis) by
z = r ei ;
z
i
for z = re , with 0 2 . Then we let f (z ) =
and consider the integral of f over the
(1 + z 2 )2
same \key-hole" contour used in problem 01.
I=
Z
Again, as R ! 1 and ! 0 the only contributions that survive are those from the straight line
pieces in . Thus we have:
lim lim
R!1 !0
Z
1
Z 0
x
(xe2i )
dx
+
dx
0 (x2 + 1)2
1 (x2 + 1)2
Z 1
x
= (1 e2i )
dx
0 (x2 + 1)2
= (1 e2i ) I :
z
dz =
(z 2 + 1)2
Z
On the other hand, the contour encloses singularities (poles) at z = i. Thus, from the Residue
theorem:
1
Z
0
f (x)dx =
=
=
=
=
=
=
2i
(Res(f; z = i) + Res(f; z = i))
1 e2i "
!
!#
2i
d
z
d
z
lim
+ zlim
! i dz (z i)2
1 e2i z!i dz (z + i)2
"
#
2i
(z + i)z 1 2z (z i)z 1 2z lim
+ zlim
! i
1 e2i z!i
(z + i)3
(z i)3
h
i
i=2 + ei=2 + e3i=2 e3i=2
e
2(1 e2i )
e i=2 ei=2
(1 ) i i
2
e
e
sin(=2)
(1 )
2
sin()
1 :
4 cos(=2)
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
1.9
Answers to Problem Set # 9.
9
Problem 06 in section 6.7.
We want to show that the polynomial p(z ) = z 6 + 4z 2 1 has exactly two roots inside the
circle jz j = 1. To this end, we use Rouche's Theorem (Theorem 4, section 6.7, page 293 in the
book). Let f (z ) = 4z 2 1 and h(z ) = z 6 . Then, on jz j = 1
jf (z)j 4jz2 j
1 = 3 > 1 = jz 6 j = jh(z )j :
Since f and h are clearly analytic for jz j < 1, the hypothesis of the theorem are satised. Hence
1
p = f + h and f have the same number of roots in jz j < 1. Since the roots of f are z = , this
2
proves the result.
Note: it need not be that p has two distinct roots in jz j < 1; it could have a single (double) root.
We note that the split above is not unique. We could take f (z ) = z 6 + 4z 2 = z 2 (z 4 + 4) and
h(z ) = 1. This works too, because f has a zero of order 2 at the origin and 4 more zeros, all
p
of which satisfy jz j = 2 (so they lie outside the unit disc). Furthermore: on the unit circle jz j = 1,
we have:
jf (z)j = jz6 + 4z2 j = jz2 jjz4 + 4j = jz4 + 4j jjzj4 4j = 3 > 1 = jh(z)j :
1.10
Problem 10 in section 6.7.
We want to show that the equation p(z ) = 2 z e z = 0 has one root in the right half
plane. Consider the functions f (z ) = 2 z and h(z ) = e z . The function f (z ) clearly has one
root in the right half plane (at z = 2). Furthermore, on the imaginary axis (write z = iy, with
1 < y < 1) we have:
jf (z)j = j2 iyj =
q
22 + y 2 2 > 1 = j
e
iy
j = jh(z)j :
Thus f (z ) and p = f + h must have the same number of zeros in the right half plane (namely 1),
by Rouche's Theorem.
Note that if z0 is a root of z = 2 e z , then z0 must also be a root of the equation. But if z0 is
in the right half plane, then so is z0 . Since we know that there is only one root in this region, we
conclude that z0 = z0 , i.e.: the root is real.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
1.11
Answers to Problem Set # 9.
10
Problem 04 in section 7.1.
We consider here the problem of nding a function that is harmonic in the right hand
plane and takes the values
y
(0; y ) =
(1.11.1)
1 + y2
z
z
on the imaginary axes. The obvious guess: = Im(
),
does
not
work
because
is not
1 z2
1 z2
analytic at z = 1. A subtler strategy is called for.
Consider the correspondence between the right half plane in the z -plane and the unit disk in the
w-plane given by the map from z to w (and its inverse):
w(z ) =
1+w
z 1
and z (w) =
:
z+1
1 w
(1.11.2)
Then (w) = (z (w)) denes a function that is harmonic on the unit disk jwj < 1, with values
on the boundary w = ei which we can calculate via the correspondence (1.11.2) and (1.11.1). Let
us now calculate these values.
(a) The correspondence between the unit disk and right hand plane boundaries is given by
z (ei ) =
1 + ei (1 + ei )(1 e i )
sin =
=
i
= i y () :
1 ei (1 ei )(1 e i )
1 cos Thus, we have
(ei )
!
sin 1
sin sin = (0; y ()) = 0;
=
:
=
2
2
1 cos 1 cos 1 + (sin =(1 cos ) )
2
1
1
1
1
2
2
2
2
Thus we have an harmonic function that satises the correct boundary conditions. Since the
solution of Laplace's equation with Dirichlet boundary conditions is unique, it follows that
this must be the one and only correct solution. That is to say, it must be:
(b) Since f (w) = w is an entire function, Im( w) is harmonic. Furthermore: Im( ei ) = sin .
1
= Im( w) :
2
(1.11.3)
(c) To nd , now we take and use the mappings (1.11.2). That is:
!
!
1
z 1
1
x + iy 1
y
(x; y ) = (w(z )) = Im( w(z )) = Im
= Im
= 2
:
2
2(z + 1)
2
x + iy + 1
y + (x + 1)2
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
1.12
Answers to Problem Set # 9.
11
Problem 07 in section 7.2.
1
Consider the mapping w(z ) = z + . An element of the circle jz j = , say z = ei , gets mapped
z
to
!
!
1 i
1
1
i
i
w(e ) = e + e = + cos + i + sin :
!
!
1
1
So we have u = Re(w) = + cos and v = Im(w) = sin . Thus:
u2
v2
+
= cos2 + sin2 = 1 :
( + 1 )2 ( 1 )2
1.13
Problem 13 in section 7.2.
Here we consider the images of various sets in the complex plane by the map
w = cos(z ) = cos(x) cosh(y ) i sin(x) sinh(y ) ;
where3 z = x + iy and w = u + iv . Thus u = cos(x) cosh(y ) and v =
sin(x) sinh(y ).
Remark 1.13.1 Analytic maps preserve orientation. Thus, when mapping boundaries of regions,
it is useful to follow the boundary keeping the region (say) on the right. Then the image of the
region will be on the right of the image of the boundary.
We follow this strategy below: the sense in which any given curve is traversed is indicated by the
1 < y 0, we mean that the line
is to be followed in the direction of increasing y. On the other hand x = 0 and 0 y > 1 means
order of the inequalities. Thus, if we write a line as x = 0 and
that the line is to be followed in the direction of decreasing y. The same strategy is used for the
image curves.
To better understand the arguments that follow below, you should draw the various sets and put
arrows on their boundaries (pointing in the direction such that the set is to the right). Then do the
same for the images of the boundaries.
(a) Half vertical strip a : 0 < x < and y < 0.
Lets look rst at what happens on the boundary:
3 That
is: x = Re(z ), y = Im(z ), u = Re(w) and v = Im(w).
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
Answers to Problem Set # 9.
12
1 < y 0.
Mapped to the half real axis: 1 > u 1 and v = 0.
This because w = cosh(y ) and 1 < y 0.
Top side: y = 0 and 0 x .
Mapped to the real interval: 1 u 1 and v = 0.
This because w = cos(x) and 0 x .
Right side: x = and 0 y > 1.
Mapped to the half real axis: 1 u > 1 and v = 0.
This because w = cosh(y ) and 0 y > 1.
Left side: x = 0 and
Thus we see that the boundary of a gets mapped to the entire real axis. Since the image
of a must be to the right of the image of the boundary (when the boundary is traversed
keeping a to the right), we conclude that:
The image of a is the upper half plane.
As a further check: consider a point z0 2 a . Then sin(x0 ) > 0 and sinh(y0 ) < 0. It follows
that: v0 = sin(x0 ) sinh(y0 ) > 0.
1
2
(b) Half vertical strip b : 0 < x < and y > 0.
Lets look rst at what happens on the boundary:
Right side: x = 12 and 1 > y 0.
Mapped to the half imaginary axis: 1 < v 0 and u = 0.
This because w = i sinh(y ) and 1 > y 0.
Bottom side: y = 0 and 12 x 0.
Mapped to the real interval: 0 u 1 and v = 0.
1
This because w = cos(x) and x 0.
2
Left side: x = 0 and 0 y < 1.
Mapped to the half real axis: 1 u < 1 and v = 0.
This because w = cosh(y ) and 0 y < 1.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
Answers to Problem Set # 9.
13
Thus we see that the boundary of b gets mapped to the lower half of the imaginary axis,
followed by the positive real axis. Since the image of b must be to the right of the image of
the boundary (when the boundary is traversed keeping b to the right), we conclude that:
The image of b is the fourth quadrant in the complex plane.
As a further check: consider a point z0 2 b . Then sin(x0 ) > 0, cos(x0 ) > 0, sinh(y0 ) > 0 and
cosh(y0 ) > 0. It follows that: u0 = cos(x0 ) cosh(y0 ) > 0 and v0 = sin(x0 ) sinh(y0 ) < 0.
(c) Vertical strip c : 0 < x < .
Lets look rst at what happens on the boundary:
1 < y < 1 (i.e.: the imaginary axis).
Mapped to the half real axis: u 1 and v = 0.
This because w = cosh(y ) and 1 < y < 1. However, watch out here: notice that
the image is traversed twice. As y goes from 1 to 1, u goes from 1 to 1 and
then back to 1 (think of the implication of this, given that the image set is to the right
Left side: x = 0 and
of the image boundary).
Right side: x = and 1 > y > 1.
Mapped to the half real axis: u 1 and v = 0.
This because w = cosh(y ) and 1 > y > 1. Again: the image is traversed twice.
As y goes from 1 to 1, u goes from 1 to 1 and then back to 1.
Thus we see that the boundary of c gets mapped to the real axis, except for the segment
1 u 1. Each of the two pieces corresponds to one of the sides of c and each is covered
twice as the boundary of c is followed (in the fashion explained above). Since the image of
c must be to the right of the image of the boundary(when the boundary is traversed keeping
c to the right), we conclude that:
The image of c is the entire complex plane, except for the
segments on the real axis given by 1 < u 1 and 1 u < 1.
1
For example: the point z = 2 c is mapped to w = 0.
2
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
Answers to Problem Set # 9.
14
(d) Rectangle d : 0 < x < and 1 < y < 1.
Lets look rst at what happens on the boundary:
Left side: x = 0 and 1 y 1.
Mapped to the real interval: 1 u cosh(1) and v = 0.
This because w = cosh(y ) and 1 y 1. But, watch out, the image is traversed
twice. As y goes from 1 to 1, u goes from cosh(1) to 1 and then back to cosh(1).
Top side: 0 x and y = 1.
Mapped to the lower half of the ellipse centered at the origin, with semi-axes cosh(1) in
the real direction and sinh(1) in the imaginary direction. The motion along it is clockwise.
This because w = a cos(x) ib sin(x), where a = cosh(1), b = sinh(1) and 0 x .
Thus u = a cos(x) and v = b sin(x), which is a parametric form of an ellipse.4
Right side: x = and 1 y 1.
Mapped to the real interval: cosh(1) u 1 and v = 0.
This because w = cosh(y ) and 1 y 1. Again, the image is traversed twice.
As y goes from 1 to 1, u goes from cosh(1) to 1 and then back to cosh(1).
Bottom side: x 0 and y = 1.
Mapped to the upper half of the ellipse centered at the origin, with semi-axes cosh(1) in
the real direction and sinh(1) in the imaginary direction. The motion along it is clockwise.
This because w = a cos(x) + ib sin(x), where a = cosh(1), b = sinh(1) and x 0.
Thus u = a cos(x) and v = b sin(x), which is a parametric form of an ellipse.5
From the above, in the same fashion as in the other cases, we conclude that:
2
2
u
v
The image of d is the elliptical set
+
< 1,
cosh(1)
sinh(1)
except for the segments on the real axis given by v = 0 and
cosh(1) < u 1 or 1 u < cosh(1).
!
4 Note
that (u=a)2 + (v=b)2 = 1. Also, v 0, since 0 x .
5 Note
that (u=a)2 + (v=b)2 = 1. Also, v 0, since 0 x .
!
Answers to Problem Set # 9.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
15
1
For example: the point z = 2 c is mapped to the origin.
2
1.14
Problem 05 in section 7.3.
The general Mobius transformation takes the form
w = u + iv =
az + b
;
cz + d
where a, b, c and d are (complex) constants. In this problem we want to nd a Mobius transformation
that maps the unit disk jz j < 1 onto the right hand plane u > 0 and takes z = i to w = 0.
The condition w( i) = 0 means that we must have a( i) + b = 0, that is b = ia .
One of the points in the unit circle must go to the point at innity (the unit disk must be
mapped to the boundary of the right hand plane; that is: the imaginary axis). We are free
to choose this point (the only point we cannot choose is z = i). Thus select z = 1 for this
purpose. Then we need c 1 + d = 0, that is d = c .
Thus far we have
a z+i
:
c z 1
Now, we need to make sure that all the points on the unit circle are mapped to the imaginary axis.
Since Mobius transformations take circles into circles,6 we only need to make sure three points on
the unit circle get mapped to the imaginary axis. We already have two; to get the third let us make
sure that z = 1 maps to a point on the imaginary axis. Thus, we must have:
w=
!
a 1 + i pa p1
1
a
w( 1) =
=
i p = p e i=4 = i;
c
2
2c
2
2
2c
p
a
for some real constant . So, choose = ei3=4 (which corresponds to = 1= 2), to obtain
c
z+i
w = e(i3=4)
:
z 1
This maps the unit circle onto the imaginary axis, but it need not map the inside of the unit circle
to the right hand plane (it could map it to the left hand plane). However, w(0) = ei=4 which is
indeed in the right half plane, and so we have a correct map.
6 Straight
lines are the special case of a circle through innity.
18.04 MIT, Fall 1999 (Rosales, Schlittgen and Zhang).
Answers to Problem Set # 9.
16
Remark 1.14.1 Notice that there are two basic ways of guring out what the image of a set (by
an analytic function) is. These are:
See where the boundary of the set gets mapped, including orientation (say, go around the set
keeping the set on the right and see which way the mapped curve is traversed). Since analytic
maps preserve orientation, the image set will be on the same side of the image boundary as
the set was relative to its boundary.
Map the boundary without concern for orientation and then check were an arbitrary point
inside the set gets mapped to. This will tell you which side of the image of the boundary the
image of the set is on. This is more eÆcient (but less intuitive).
THE END.